77-1 Spur Gear

The following data is given for a pair of spur gears with

20

 full depth involute teeth.

Number of teeth on pinion= 24, Number of teeth on gear= 56, Speed of pinion= 200 rpm, !odule= "mm, #a$e width= "0mm. %oth the gears are made of o f steel with an ultimate tensile strength of, 600 N&mm2. 'sing the velo$it( fa$tor to a$$ount for the d(nami$ load, $al$ulate ) %eam strength* 2) +elo$it( fa$tor* and ") ated power that the gears $an transmit without bending failure, if the fa$tor of safet( is .5. Solutionata givenφ 

=

20 

full depth involutes,

 Number of teeth on pinion / p) = 24,  Number of teeth on gear /g) = 56,  Speed of pinion /n p) = 200 rpm,  !odule /m) = "mm,  #a$e width /b) = "0 mm, Servi$e fa$tor /1s) = .5,  N 

'ltimate tensile strength /Sut) = 600

mm

2

.

s both gear and pinion are made of same material, material, then pinion is a vehi$le vehi$le member. So the design is based on pinion. d  p

/ m) =

 z  p

=

d  g   z  g 

3et, module d  p

 

So,

d  g  = m.z  g 

= m.z  p

×

= " 24 = 2mm

×

= " 56 = 6mm /) %eam stre strengt ngth h /S b)-

S b

= σ  b × m × b × Y 

( A) ..

σ  b

s

∴

=

S ut 

"

20

=

600 "

= 200 N 

mm

2

full depth involute s(stem,

    

Y  = π   0.54 −

0.72    

   

 z  p

0.72   = π     0.54 −   24    

= π  ( 0.54 − 0.6") = π   × 0.6 ∴ Y  = 0."64

= σ  b × m × b × Y 

S b So, beam strength = 200 × " × "0 × 0."64 S b

= 640 N C v

/2) +elo$it( fa$tor /

)-

( v) = 8it$h line velo$it(

= v

d  n p

π    p

60 × 000

× 2 ×200 60 ×000

π  

= 4.524 m s v

s the pit$h line velo$it( / ) 9 0 C v So, +elo$it( fa$tor,

=

" "+v

m  s

,

=

" " + 4.524

C v

=

 0."7

/") 8ower /8)S b

=  peff  . f   s

%eam strength, / p eff   ) =

S b  f   s

=

640 .5

So, :ffe$tive load, = 4"20 N 2π  n pT  / P ) = ( B ) 60 × 000 3et, 8ower  / peff  ) =

C  s C v

pt 

:ffe$tive load  pt 

 

=

C v C  s

So, 0."7 × 4"20 .5  pt 

Tangential load,

 pt  =

= 4.26 N 

2T  d  p

;ere, T  =

Therefore,

=

=

 pt d  p

2

4.26 × 2 2

T  = 4"".26 N .mm

#rom e4  pair of spur gear is to be designed for a ma$hine operating at 240rpm, driven b( an ?@, 200rpm ele$tri$ motor. 1entre distan$e between the aAis of the gear shaft ma( be approAimatel( "00mm. Starting tor 0.912

 z

, where  is the number of teeth on gear.

SD3'TEDN!a$hine rpm= 240 :le$tri$ motor rpm= 200 Speed ratio= 5 Sa(, d p = pit$h $ir$le dia. Df pinion

d p =

Then, pit$h $ir$le dia. Df gear = 5 d p+d 1enter distan$e= "00mm= Dr, 600= d p F Dr,

g

2

d g = d p F5 d p

d p =00mm

d g  =500mm

dg

2 π × 1200

Speed, +=

60

×

 0.1 2

 = 6.2" m&s

3

+elo$it( fa$tor, 1v =

3+ v

3

 =

3 + 6.283

 = 0."2"

startingtorque rated torque  = .5

Servi$e fa$tor, 1s =

 KW × 1000 8t, tangential tooth load = v 8000

=

6.283

 = 2".2 N

Cs :ffe$tive load,  peff   = 8t Cv %eam strength,

Sb

1.5

= 2".2G

0.323

 = 57" N

= mb σ b H

#a$e width, b = 0m @here, m is module 3ewis form fa$tor H= I( =I

(

0.154 −

0.912

 z

)

#a$tor of safet( = 2

S b =  peff  G#DS = 57"G2 = 26 N σ b =

σ ut  3

600

 =

3

26 = bmH σ b m

2

 = 200!pa

= 0mG200GmGH

(=5.7"

!odule of 4mm $an be safel( $hosen for pinion and gear. d p = 4G25=00mm

d g = 4G25=500mm

;N:SS TaBing

Sw =

S b = 26N

b=0m= 40mm d p =00mm 2 z g

atio fa$tor, J = 2 × 125 125 × 25

−

=

 z g + z b

250

=

150

.66

S b = b d p J?  26= 40G00G.66G?  11826

3oad stress fa$tor, ?=

4000 × 1.667

= ." = 0.6

(  ) BHN 

2

100

%;N = """ Surfa$e hardness is """ %;N

 pair of straight teeth spur gears, having 20K involutes full depth teeth is to transmit 2 Bw at "00r.p.m of the pinion. The speed ratio is "-. The allowable stati$ stresses for gear of $ast iron and pinion of steel are

ln ❑

 60!8a and 05!8a respe$tivel(. ssume the following,

 Number of teeth on pinion =6 #a$e width =4times module +elo$it( fa$tor/ c v ¿ =

4.5 4.5 + v

−0.912

fa$tor H=0.54 noof teet! .

 ,   v being the pit$h the line velo$it( in m&s-and tooth form

etermine the module fa$e width and pit$h diameter of gears, $he$B the gears for wear for  2 2 wear, given σ =600 "#a $ ϵ  p =200 %N / mm ∧ϵ g =100 %N / mm ,sBet$h the gears.

SolutionLiven data-

∅=20

2

σ '( =60  N / mm

&

2

3

 

σ '# =105 N / mm

× 10 W 

8=2B@=2

 N  p=300 r  p  m

)  #=16

) (

+.= )  =3  p

 

b =14 m 3

2

2

 * p= 200 × 10  N / mm 3

σ CS =600 N / mm

2

 * g=100 × 10  N / mm

!odule @e Bnow that pit$h line velo$it(,

+ =

π  p  N  # 60

 @e taBe from the table W ) =

=

πm× 16 × 300 60

=251 mm / s = 0.251 m / s

C S=1

 # ×C S +  3

=

= 4.5

+=

4.5 + v

=

12 × 10

0.251 m

47.8 × 10

 × 1

3

m

N 

4.5 4.5 + 0.251 m

@e Bnow that form fa$tor for pinion and gear,

and velo$it( fa$tor 

 - p=0.154 −

 - g =0.154 −

0.912

) g

0.912

) (

=0.154 −

= 0.154−

0.912 16

0.912 3 × 16

=0.097

= 0.135

σ '# × - p =105 × 0.097 =10.085 nd Sin$e

σ C( × . (= 60 × 0.135= 8.1

( σ 

×. ( )

C(

 is less than / σ C( × . ( ¿ .there for the gear is weaBer, now using 3ewis

em

) ma1  = .25T . !odule and fa$e width of the gears m = module in mm, and  b = fa$e width in mm. sin$e the starting tor